Question

Find tan θ if sin θ = 3/4 and θ terminates in QI

Answer 1

`\bf sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{4}}\impliedby \textit{now let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm√(c^2-b^2)=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`


`\bf \pm√(4^2-3^2)=a\implies \pm√(16-9)=a \\\\\\ \pm√(7)=a\implies \stackrel{\textit{in the I Quadrant}}{+√(7)=a}\qquad therefore \\\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{3}{√(7)} \\\\\\ \textit{and rationalizing the denominator}\qquad tan(\theta)=\cfrac{3√(7)}{7}`